Below is test of ten million trials to see when a repeat occurs ...

[ego]

Below is test of ten million trials to see when a repeat occurs
on a double zero roulette table using a random number generator.

Spin: Repeat occurred
1  : 0
2  : 263280
3  : 512608
4  : 729706
5  : 893503
6  : 1000672
7  : 1041586
8  : 1023667
9  : 955786
10 : 848365
11 : 718483
12 : 583348
13 : 450958
14 : 334641
15 : 237568
16 : 160392
17 : 103437
18 : 63660
19 : 37204
20 : 20665
21 : 10767
22 : 5381
23 : 2531
24 : 1067
25 : 444
26 : 173
27 : 78
28 : 26
29 : 3
30 : 0
31 : 0
32 : 0
33 : 0
34 : 0
35 : 0
36 : 0
37 : 0
38 : 0
39 : 0
40 : 0
41 : 0
42 : 0
43 : 0
44 : 0
45 : 0
46 : 0
47 : 0
48 : 0
49 : 0

[superman]

Hi ego

When you say repeat do you refer to 12,12 or 12,29,34,12?

thanks

[Drazen]

Thanks  Ego and we know that most repeating is on 7th spin. But can you simulate same thing with 0 roulette and of course to tells us maybe most important thing. How much zero spuns in that number of spins on 7th spin ?

Regards

Drazen

[Juiced91]

I would assume that it is 1 after the other so, 6 6. or 2 2.

I was made a system to follow repeaters and bet on a number to come up again. so if 4 came i bet 4 again. But there was sometimes 350 spins without a repeat. It worked for long before giving in

[superman]

But there was sometimes 350 spins without a repeat.

Yes my reply would have been something like that too, that's the problem even on BV NZ or european wheels.

Before drazen gets upset, we know RNG is not roulette although casinos call it roulette, maybe we could sue for false advertising!!

[Bayes]

I would assume that it is 1 after the other so, 6 6. or 2 2.

I don't think that's right. I generated the same table for single 0 and posted it here. But I can't remember what the numbers actually mean because it was over a year ago. 

I'm pretty sure it wasn't a repeat in consecutive spins though, because the average wait for that event is 37 spins.

[Juiced91]

I don't think that's right. I generated the same table for single 0 and posted it here. But I can't remember what the numbers actually mean because it was over a year ago. 

I'm pretty sure it wasn't a repeat in consecutive spins though, because the average wait for that event is 37 spins.

Well i have waited for over 350 spins for ANY repeat

[mr.ore]

Similar result to bayes's for one million spins:

1 3205
2 6270
3 8866
4 10975
5 12295
6 12764
7 12597
8 11484
9 10284
10 8558
11 6964
12 5161
13 3852
14 2713
15 1763
16 1107
17 652
18 396
19 231
20 81
21 53
22 22
23 15
24 7
25 3
26 0
27 0
28 1
29 0
30 0
31 0
...

The most common spin to see ANY repeater is sixth spin. The repeater can be any number already spun. So one experiment looks like 13 21 14 17 13 and the result would be 4 spin because 13 repeated after four spins. Then reset all and measure again, throwing history - that means more spins are needed. EDIT: I used spins, not trials to measure... i'm really retarded.

[mr.ore]

My version over one hundred million spins (not trials):

0 0
1 326660
2 633199
3 897713
4 1100517
5 1228009
6 1272982
7 1244513
8 1154504
9 1017087
10 855385
11 686307
12 526720
13 385202
14 269496
15 178703
16 113599
17 68085
18 39230
19 21144
20 10963
21 5340
22 2444
23 1068
24 403
25 143
26 52
27 20
28 5
29 0
30 0
31 0

[ophis]

I'm sorry but where did you get those sample spins from? (1kkk and 10kkk)

could you share them with us? I would like to verify even distribution in them just for sake of things

or was that just coded RNG?

[mr.ore]

Just RNG...

link:://:.gnu.org/s/gsl/manual/html_node/Random-number-generator-algorithms.html

Generator: gsl_rng_ranlux389
The ranlux generator is an implementation of the original algorithm developed by Lüscher.  It uses a lagged-fibonacci-with-skipping algorithm to produce “luxury random numbers”.  It is a 24-bit generator, originally designed for single-precision IEEE floating point numbers.  This implementation is based on integer arithmetic, while the second-generation versions ranlxs and ranlxd described above provide floating-point implementations which will be faster on many platforms. The period of the generator is about 10^171.  The algorithm has mathematically proven properties and it can provide truly decorrelated numbers at a known level of randomness.  The default level of decorrelation recommended by Lüscher is provided by gsl_rng_ranlux, while gsl_rng_ranlux389 gives the highest level of randomness, with all 24 bits decorrelated. Both types of generator use 24 words of state per generator.
For more information see,

   
• M. Lüscher, “A portable high-quality random number generator for lattice field theory calculations”, Computer Physics Communications, 79 (1994) 100â€"110.
• F. James, “RANLUX: A Fortran implementation of the high-quality pseudo-random number generator of Lüscher”, Computer Physics Communications, 79 (1994) 111â€"114

[ophis]

Thanks. In this case I think there is noting to validate. 

[MrJ]

A few years back (at VIP), we had a pretty decent method based on this. (I already knew it was around the 6th-7th spin)

All flat betting (2 numbers), bet on the 3rd and 9th number on the history board. If by chance they are both the same, double up on that number.

Ken

[ego]

If you take one repeat of the past or previous five numbers with no repeat - then you have the ultimate window for attack.
Previous five has the highest probability due two factors.

First one is due to that the simulations show that one attack window of three attempts is ultimate to catch one repeat among the past previous five with no repeat.
Second if you where lucky to catch a bias wheel you would reduce the house edge and catch does favorites that would be among the previous five.

I read many topics regarding this about repeats and as Snowman puts it - even if there is no bias you reduce the house edge to certain degree - but does not make it vanish on certain real wheels.
That is good sign and make it worth looking deeper into the topic.

It also remained me of one similar topic made by Bayes at VLS and i quote:

This is a short review of a book called "The Fortune Maker Tackles Roulette" (available on Amazon @ $13.99).

You'll need to buy the book to see the details fleshed out, but the basic idea, and the heart of the method, is the following:

Collect some spins, the author recommends at least 1000 from each wheel, and for each number:

calculate how much you would have won (for all spins) betting the previous number and its neighbours - for 2 neighbours, then another 2, then another 2 etc, so you are doing the calculations for betting on  3,5,7,11,13,15,17,19,21 numbers. e.g. if your spin sequence from a wheel started - 23,4,13,29... you would compute your winnings betting 3,5,7 etc numbers centred around 23. Number 4 came up next, which would have been a loss for all, because even betting 21 numbers this would encompass a sector from 25 through to 9, which doesn't include number 4.

Having done all this, the next step is to make a comparison between the amount of numbers you would have bet, in terms of how much each would have won. The amount of numbers which gave the best performance will be the amount of numbers to bet (the last number spun and its neighbours). You then play that wheel using those numbers.

Obviously, to make a direct comparison, you need to "equalize" the results, because if you're betting 15 numbers then this is going to cost you more than if only betting 5. ie; 5 numbers @ $1 per number = $5, and 15 numbers @ $1 per number = $15. So in order to compare the two, you need to raise the bet level of the 5 number strategy to $3 per number ($3 × 5 = $15).

The multiplier is actually referenced to the maximum number of number you would play ie; 21.

So,

21 numbers @ $1 = $21
19 numbers @ $1.1053 = $21
17 numbers @ $1.2353 = $21
15 numbers @ $1.4000 = $21

etc.

Do this for each number you've collected, and record the results. You should then be able to determine, on the basis of a direct comparison, which (if any) number strategy is best for that wheel - whether 3,5,7 etc numbers should be played.

The idea is that IF any kind of bias is present, this analysis will pick it up. It is recommended that even if you don't collect 1000 numbers, you should always do what he calls a "dry run" before playing any wheel. This means collecting 37 numbers (one "cycle") and doing the above analysis. If any number strategy stands out as being particularly profitable, then by all means play it.

The author emphasises that it may not be the case that you have found any bias even if a strategy is suggested by the data generated by the dry run (37 spins is too small a sample) - it may just be "luck". However, there are also detailed explanations of how to closely monitor your results while playing, so you know when to quit.

It's not necessary, the author maintains, to know the cause of the bias, and states that:
He defines the effect of a bias as:
And this provides the rationale for the method described.

So that's basically it. Obviously, to do all these calculations by hand would be incredibly time-consuming and tedious, so you'll need to set up a spreadsheet or write a computer program. The author gives a url where you can download tools, but this seems to no longer exist.

The author also recommends studying the data to try to find other kinds of bias which are not necessarily keyed to the last number played. For example, you might notice that the number 5 has hit twice in a cycle (37 spins), and it's conceivable that the ball tried to get into this pocket on other occasions but didn't quite make it. Examining the data might reveal this to be so, in which case you might decide to target 5 and surrounding numbers. The above scenario might have occurred for more than one number, so you could end up playing more than one sector.

It would have been better if there were some guidelines on how to find statistical significance in the data, instead of merely telling you to pick the sector which won the most, which isn't very rigorous. Also, if you can't determine the cause of the bias, you're on shaky ground, but this isn't easy. Presumably, if you can do it, so can the casinos, that's why they have regular maintenance schedules.

However, in spite of these limitations,  the book is worth reading. There are chapters on money management and detailed charts showing, for each number strategy (whether playing 3,5,7 etc) whether you are in the "win zone" and if so what is the strength of the bias.

Question:

If 678 after previos five with no repeat is the ultimate then what happens if i do the following - does it change probability to get long losing run?

If i pick five high and low previos number with no repeat and follow high or low - then i should have two horses in favour!!!
One that the previos high or low will become a serie of six or i should have one repeat within three attempts.

Then would you not agree with this senario i would be better of follow the previos five then playing against them?

Just run for fun some short sampels.
13 attacks where 8 include one repeat and all won.

1
5
4
7
8 Previos 5 Low

3 Won
22
35

- - -

20
29
33
31
26 Previos 5 High

7  Loss
32 Won
26 Repeat

- - -

28
19
34
24
25 Previos 5 High

32 Won
1
24 Repeat

- - -

26
28
31
27
23 Previos 5 High

20 Won
25
26 Repeat

- - -

35
20
29
32
23 Previos 5 High

3  Loss
27 Won
17

- - -

28
33
20
35
29 Previos 5 High

30 Won
1
15

- - -

9
11
1
7
18 Previos 5 Low

11 Won Repeat
9  Repeat
9  Repeat

- - -

30
24
26
21
27 Previos 5 High

28 Won
25
32

- - -

22
29
30
19
25 Previos 5 High

11 Loss
14 Loss
21 Won

- - -

24
28
21
22
31 Previos 5 High

0  Loss
36 Won
32

- - -

22
27
20
25
29 Previos 5 High

27 Won Repeat
11
23

- - -

24
23
33
22
31 Previos 5 High

29 Won
36
35

- - -

10
17
12
5
18 Previos 5 Low

17 Won Repeat
20
31

[Juiced91]

If you take one repeat of the past or previous five numbers with no repeat - then you have the ultimate window for attack.
Previous five has the highest probability due two factors.

First one is due to that the simulations show that one attack window of three attempts is ultimate to catch one repeat among the past previous five with no repeat.
Second if you where lucky to catch a bias wheel you would reduce the house edge and catch does favorites that would be among the previous five.

I read many topics regarding this about repeats and as Snowman puts it - even if there is no bias you reduce the house edge to certain degree - but does not make it vanish on certain real wheels.
That is good sign and make it worth looking deeper into the topic.

It also remained me of one similar topic made by Bayes at VLS and i quote:

Question:

If 678 after previos five with no repeat is the ultimate then what happens if i do the following - does it change probability to get long losing run?

If i pick five high and low previos number with no repeat and follow high or low - then i should have two horses in favour!!!
One that the previos high or low will become a serie of six or i should have one repeat within three attempts.

Then would you not agree with this senario i would be better of follow the previos five then playing against them?

Just run for fun some short sampels.
13 attacks where 8 include one repeat and all won.

1
5
4
7
8 Previos 5 Low

3 Won
22
35

- - -

20
29
33
31
26 Previos 5 High

7  Loss
32 Won
26 Repeat

- - -

28
19
34
24
25 Previos 5 High

32 Won
1
24 Repeat

- - -

26
28
31
27
23 Previos 5 High

20 Won
25
26 Repeat

- - -

35
20
29
32
23 Previos 5 High

3  Loss
27 Won
17

- - -

28
33
20
35
29 Previos 5 High

30 Won
1
15

- - -

9
11
1
7
18 Previos 5 Low

11 Won Repeat
9  Repeat
9  Repeat

- - -

30
24
26
21
27 Previos 5 High

28 Won
25
32

- - -

22
29
30
19
25 Previos 5 High

11 Loss
14 Loss
21 Won

- - -

24
28
21
22
31 Previos 5 High

0  Loss
36 Won
32

- - -

22
27
20
25
29 Previos 5 High

27 Won Repeat
11
23

- - -

24
23
33
22
31 Previos 5 High

29 Won
36
35

- - -

10
17
12
5
18 Previos 5 Low

17 Won Repeat
20
31

Busy running tests on this idea. that 5 numbers will repeat in 6 spins..